Question

find the Wronskian of the given pair of functions. $$e^t\sin t,e^t\cos t$$

Short answer

Question: Determine the Wronskian of the given pair of functions: $$f_1(t)=e^t\sin t$$ and $$f_2(t)=e^t\cos t$$. Answer: The Wronskian of the given pair of functions is $$W(f_1,f_2)(t)=e^{2t}({\cos }^{3}t+{\cos }^{2}t\sin t-1)$$.

Step-by-step solution

Find the derivatives of the functions

Find the first derivative of each function with respect to t. The first function is: $$f_1(t)=e^t\sin t$$ Applying the product rule, we get: $$f_1^{\prime }(t)=\frac{d}{dt}(e^t\sin t)=e^t\sin t+e^t\cos t$$ Similarly, the second function is: $$f_2(t)=e^t\cos t$$ And its first derivative is: $$f_2^{\prime }(t)=\frac{d}{dt}(e^t\cos t)=e^t\cos t-e^t\sin t$$

Calculate the Wronskian

The Wronskian W(f_1, f_2) of the given pair of functions is the determinant of the following matrix: $$W(f_1,f_2)(t)=\left|\begin{array}{cc}{f}_1(t)& f_2(t)\\ f_1^{\prime }(t)& f_2^{\prime }(t)\end{array}\right|$$ Substitute the functions and their derivatives into the matrix: $$W(f_1,f_2)(t)=\left|\begin{array}{cc}{e}^t\sin t& e^t\cos t\\ e^t\sin t+e^t\cos t& e^t\cos t-e^t\sin t\end{array}\right|$$

Evaluate the determinant and simplify

Now, we'll evaluate the determinant and simplify the result: $$W(f_1,f_2)(t)=(e^t\sin t)(e^t\cos t-e^t\sin t)-(e^t\cos t)(e^t\sin t+e^t\cos t)$$ Factor out $e^{2t}$: $$W(f_1,f_2)(t)=e^{2t}(\sin t\cos t-{\sin }^{2}t-{\cos }^{2}t\sin t-{\cos }^{3}t)$$ Now, using the identity ${\sin }^{2}t+{\cos }^{2}t=1$, we can simplify the expression further: $$W(f_1,f_2)(t)=e^{2t}(\sin t\cos t-1+{\cos }^{2}t\sin t+{\cos }^{3}t)$$ And finally, combining the terms, we get: $$W(f_1,f_2)(t)=e^{2t}({\cos }^{3}t+{\cos }^{2}t\sin t-1)$$ So, the Wronskian of the given pair of functions is: $$W(f_1,f_2)(t)=e^{2t}({\cos }^{3}t+{\cos }^{2}t\sin t-1)$$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a central role in engineering, physics, economics, and other sciences, serving as the cornerstone in modeling the behavior of various systems. For example, they can describe the growth of populations, the motion of particles, and the change in investment returns over time. Learning to solve differential equations not only equips one with techniques to tackle specific problems but also develops a deeper understanding of how changes in a system lead to observable outcomes.

Typically, there are two main types of differential equations: ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives; and partial differential equations (PDEs), dealing with functions of multiple variables and their partial derivatives. In the classroom, we frequently start by studying ODEs, progressing to more complex PDEs and systems of differential equations.

Linear Independence

The concept of linear independence is crucial when working with vectors in a vector space, and it translates expectantly well to functions in the context of differential equations. A set of functions is said to be linearly independent if no function in the set can be written as a linear combination of the others. In simpler terms, this means that each function adds new, unique information to the set, much like different chapters in a book telling a comprehensive story.

To determine if functions are linearly independent, one tool we use is the Wronskian, named after the Polish mathematician Józef Hoene-Wronski. If the Wronskian of a set of functions is not zero for at least one value of the independent variable, then the functions are considered to be linearly independent. This property is particularly important in solving systems of differential equations since it affects the structure of the solution set.

Product Rule

In calculus, the product rule is an essential technique used when taking derivatives of products of two functions. It tells us that the derivative of a product is not simply the product of the derivatives, which is a common misconception. Instead, the rule states that the derivative of the product of two functions, say f and g, is given by f'g plus fg'. Formally, if f(t) and g(t) are both differentiable, then the product rule can be written as:
$$(fg{)}^{\prime }(t)=f^{\prime }(t)g(t)+f(t)g^{\prime }(t)$$
This concept is elegantly reflected in our previous exercise, where we applied the product rule to find the derivatives of two functions involving exponential and trigonometric terms.

Determinant

The determinant is a scalar attribute of a square matrix that provides a lot of information about the matrix. For instance, the determinant can tell us if a matrix is invertible, by checking if it is nonzero. In two dimensions, it gives us the scale factor of the linear transformation described by the matrix, and the area transformation of a unit square. In the context of differential equations, the determinant is used to form the Wronskian, which helps us analyze the linear independence of solutions.

For a 2x2 matrix, the determinant is found by subtracting the product of the top-right and bottom-left entries from the product of the top-left and bottom-right entries, as shown in the previous step by step solution. Determinants can be more challenging to compute as matrix size increases, but the concept remains a pivotal part of understanding linear algebra and its applications in differential equations.